Question Video: Finding the Average Value of a Quadratic Function on a Given Interval | Nagwa Question Video: Finding the Average Value of a Quadratic Function on a Given Interval | Nagwa

# Question Video: Finding the Average Value of a Quadratic Function on a Given Interval Mathematics • Higher Education

Find the average value of π(π₯) = (2π₯ β 5)Β² on the interval [1, 4].

03:12

### Video Transcript

Find the average value of the function π of π₯ is equal to two π₯ minus five squared on the closed interval from one to four.

The question is asking us to find the average value of our function π of π₯ on the closed interval from one to four. And we recall that the average value of a function π of π₯ on the closed interval from π to π is given by π average is equal to one divided by π minus π multiplied by the integral from π to π of π of π₯ with respect to π₯. And this assumes that our function π of π₯ is continuous on the closed interval from π to π. We donβt need to worry about that in this case. Our function π of π₯ is a polynomial which is continuous on the whole real numbers.

Since we want to find the average value on the closed interval from one to four, weβll set π equal to one and π equal to four. So the average value of our function on the closed interval from one to four is equal to one divided by four minus one multiplied by the integral from one to four of two π₯ minus five squared with respect to π₯. We have four minus one is equal to three. And we might be tempted to evaluate our integral by using a substitution such as π’ is equal to two π₯ minus five, which would work. However, since weβre only squaring a linear term, itβs easier to distribute our parentheses and integrate term by term.

Distributing the exponent over our parentheses gives us four π₯ squared minus 20π₯ plus 25. To evaluate this integral, weβll use the power rule for integration. Which tells us for constants π and π, where π is not equal to negative one, the integral of ππ₯ to the πth power with respect to π₯ is equal to ππ₯ to the power of π plus one divided by π plus one plus the constant of integration πΆ. We add one to the exponent and then divide by this new exponent.

To integrate our first term, we add one to the exponent to get three and divide by this new exponent of three. This gives us four π₯ cubed over three. And since weβre calculating a definite integral, we donβt need to worry about the constant of integration. To integrate our second term, we remember that π₯ is equal to π₯ to the first power, so we add one to our exponent and divide by this new exponent. This gives us negative 20π₯ squared over two, which simplifies to negative 10π₯ squared.

Finally, we can integrate our third term by remembering that 25 is equal to 25 times π₯ to the zeroth power since π₯ to the zeroth power is equal to one. Adding one to our exponent and then dividing by this new exponent gives us 25π₯ to the first power all divided by one, which simplifies to give us 25π₯. This gives us one-third times four π₯ cubed over three minus 10π₯ squared plus 25π₯, evaluated at the limits of integration π₯ is equal to one and π₯ is equal to four.

Next, we evaluate this at the bounds of our integral, π₯ is equal to one and π₯ is equal to four. Evaluating these expressions gives us one-third multiplied by 256 over three minus 160 plus 100 minus four-thirds plus 10 minus 25, which we can calculate to give us three.

Therefore, weβve shown the average value of the function π of π₯ is equal to two π₯ minus five all squared on the closed interval from one to four is equal to three.

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